Pi

Suppose $\pi =a/b$. Define

$$f(x)=\frac{x^n(a-\mathrm{bx}{)}^n}{n!}$$

and

$$F(x)=f(x)-f^{(2)}(x)+f^{(4)}(x)-...+(-1{)}^n{f}^{(2n)}(x)$$

for every positive integer $n$.

First note that $f(x)$ and its derivatives $f^{(i)}(x)$ have integral values for $x=0$, and also for $x=\pi =a/b$ since $f(x)=f(a/b-x)$.

We have

$$\frac{d}{dx}(F\prime (x)\sin x-F(x)\cos x)=F″(x)\sin x+F(x)\sin x=f(x)\sin x$$

whence

$${\int }_0^{\pi }f(x)\sin x\mathrm{dx}=[F\prime (x)\sin x-F(x)\cos x{]}_0^{\pi }=F(\pi )+F(0)\in \mathbb{Z}$$

But for $0<x<\pi$, we have

$$0<f(x)\sin x<\frac{{\pi }^n{a}^n}{n!}$$

which means we have an integer that is positive but tends to zero as $n$ approaches infinity, which is a contradiction.